Linear amplifiers have been used for many years in amplification systems. They have numerous advantages over switch mode amplifiers, such as lower noise, lower distortion, lower EMI (Electro-Magnetic Interference), lower component count, and lower cost. However, linear amplifiers in general have lower power efficiency than switch mode amplifiers. The relatively low power efficiency of linear amplifiers may shorten battery run-times of battery-powered consumer electronic devices. In spite of the lower efficiency, currently linear amplifiers are more widely used in portable electronic devices than switch mode amplifiers because of the above-mentioned advantages.
A typical application of a linear amplifier is an audio amplifier for a consumer electronic device. In such applications, the audio amplifier amplifies a baseband signal, representing audio content, in order to drive a transducer, such as a speaker or headphone, to generate an acoustic signal. Various amplifier circuits have been used. FIG. 1 illustrates an audio amplifier employing two linear amplifiers to drive load 102 in response to input signal Vin(t). In this example, input signal Vin(t) is a baseband signal audio with a zero time-average. That is, its DC (Direct Current) component is zero, so that the time average of Vin(t) over long time periods is essentially zero. (The audio information is represented by the variations of Vin(t) about zero, or ground.) Load 102 may be, for example, a speaker or headphone. The circuit of FIG. 1 is sometimes referred to as an H-bridge amplifier due to its resemblance of the letter “H”.
In FIG. 1, input signal Vin(t) is provided to the inverting input port (or terminal) of amplifier 110, also called the negative input port (or terminal). (In other embodiments, Vin(t) may be provided to the non-inverting input port.) Amplifier 110 may be, for example, an OPAMP (Operational Amplifier). The combination of amplifier 110 and resistors R1 and R2 comprises linear amplifier 112, where resistors R1 and R2 set the gain of linear amplifier 112. Denoting the output voltage of linear amplifier 112 as Vo1(t), the gain of linear amplifier 112 is defined as δVo1(t)/δVin(t), the partial derivative of the voltage Vo1(t) with respect to the voltage Vin(t). This gain is often referred to as the AC (Alternating Current) gain. The gain of linear amplifier 112 is well known and is given as −R1/R2, where now R1 and R2 in this expression represent the resistances of resistors R1 and R2, respectively. (For ease of notation, in these letters patent a symbol used to denote a resistor in a drawing also represents its resistance value when used in an expression. That is, if a resistor is designated in a drawing by the symbol R, then its resistance value will also be indicated by the same symbol R.)
In FIG. 1, the combination of amplifier 120 and resistors R3 and R4 comprise linear amplifier 114. Resistors R3 and R4 are usually chosen to be equal in value. With this choice, the output voltage Vo2(t) of linear amplifier 114 is the negative of Vo1(t), i.e., Vo2(t)=−Vo1(t). Because of this, it is often said that the H-bridge amplifier has a differential output, and is driving its load in a differential mode. The load voltage is the voltage difference (Vo1(t)−Vo2(t)) applied to load 102, and the overall gain, Gn, for the H-bridge amplifier of FIG. 1 is Gn=δ(Vo1(t)−Vo2(t))/δVin(t)=2*δVo1(t)/δVin(t)=−2*R2/R1. Note that in FIG. 1, Vps is the positive power supply voltage and −Vps is the negative power supply voltage.
FIG. 2a is another example of an audio amplifier, but employing only one linear amplifier 204, comprising amplifier 202 and resistors R1 and R2. (For simplicity, we will re-use symbols for resistors in various drawings. This is not meant to imply that resistors in different drawings labeled with the same symbol necessarily have the same resistance. It will be clear from context with drawing is being referred to, and therefore there should be no cause for confusion.) The gain of linear amplifier 204 is given by −R2/R1, and because only one linear amplifier is employed, this gain is also the overall gain for the circuit.
Yet another amplifier circuit is shown in FIG. 2b, again employing only one linear amplifier 210, comprising amplifier 212 and resistors R1 and R2. However, unlike the circuits of FIGS. 1 and 2a, only one power supply voltage, denoted as Vps, is needed. Accordingly, the input signal Vin(t) in FIG. 2b should have its time-average equal to Vps/2, and a bias voltage Vps/2 is applied to the non-inverting input port of amplifier 212. (Although the input signal voltage in FIG. 2b has a different time-average than that shown in FIGS. 1 and 2a, for notational convenience the same symbol, Vin(t), is used. This should be no cause for confusion.) To prevent wasted static power due to this Vps/2 average value, capacitor 215 is employed in series with load 214. Capacitor 215 is often referred to as a coupling capacitor.
The output voltage, Vo(t), of amplifier 210 is given by Vo(t)=G[Vin(t)−Vps/2]+Vps/2, where the gain G is G=−R2/R1. Capacitor 215 will have a time-average voltage difference of Vps/2, so that the load voltage applied to load 214 is Vload(t)=Vo(t)−Vps/2=G[Vin(t)−Vps/2]. (Under ideal conditions, if capacitor 215 had unlimited capacitance, the voltage drop across it would be constant, equal to Vps/2. The larger the capacitance, the less its voltage drop varies with time.)
In comparing FIGS. 2a and 2b, we see that the circuit of FIG. 2b has the advantage of requiring only one power supply voltage, but at the expense of requiring a capacitor. Capacitor 215 in series with load 214 is essentially a simple high-pass filter, and will filter out the low frequency components of the output voltage. Accordingly, in audio applications, capacitor 215 may limit audio fidelity in the bass region of the output audio signal unless it is large enough to pass the low frequencies of interest. But too large of a capacitor may not be suitable for a small, portable electronic device.
The circuits of FIGS. 2a and 2b have the same power efficiency, assuming that capacitor 215 is large enough to pass most of the desired audio signal. Note that to have the same peak output power as that of the circuit in FIG. 2a, the value for the power supply voltage Vps in FIG. 2b should be two times the value of the power supply voltage used in FIG. 2a. For the remaining discussion in this Background, we assume this relationship for the values of power supply voltage Vps used in the circuits of FIGS. 2a and 2b. 
In general, a linear amplifier is linear over some range of its input voltage. As the output voltage of a linear amplifier approaches a supply rail (e.g., one of the power supply rails Vps or -Vps for the circuit of FIG. 1 or FIG. 2a, or the power supply rail Vps or ground for the circuit of FIG. 2b), the linear amplifier may saturate. For example, a simplified model for a linear amplifier requiring the two power supply voltages Vps and −Vps, such as that used in the H-bridge amplifier of FIG. 1 or used in the amplifier of FIG. 2a, is illustrated in FIG. 17a, where Vout(t) is the output voltage, Vin(t) is the input voltage, Vsat is a saturation voltage in the range 0<Vsat<Vps, and where we assume that the gain is negative. The functional relation, generalized to either a negative or positive gain G, may be written as: Vout(t)=GVin(t) for |Vin(t)|≦Vsat/|G|; Vout(t)=G*Vsat/|G| for Vin(t)>Vsat/|G|; and Vout(t)=−G*Vsat/|G| for Vin(t)≦−Vsat/|G|.
As another example of saturation, a simplified model for a linear amplifier requiring only one power supply voltage Vps, such as that used in the circuit of FIG. 2b, is illustrated in FIG. 17b, where the saturation voltage is in the range 0<Vsat<Vps/2, and where again we assume that G is negative. The functional relation, generalized to either a negative or positive gain G, is: Vout(t)=G[Vin(t)−Vps/2]+Vps/2 for |Vin(t)−Vps/2|≦Vsat/|G|; Vout(t)=G*Vsat/|G|+Vps/2 for Vin(t)−Vps/2>Vsat/|G|; and Vout(t)=−G*Vsat/|G| for Vin(t)−Vps (Vsat depends upon the power supply voltage, so that if the power supply voltage for the circuit of FIG. 2b is twice that of FIG. 2a in order for the two circuits to provide the same peak output power, then ideally the saturation voltage for each circuit will have the same value.)
With the above simple model in mind, for a single-amplifier circuit, such as the circuit of FIG. 2a or FIG. 2b, the magnitude of the voltage drop across the load, the load voltage, does not exceed Vsat. (Remember, we are assuming that the value of the supply voltage used for FIG. 2b is twice that used in FIG. 2a, so that the numerical values for their saturation voltages are the same, which we denote simply as Vsat.) However, for the H-bridge amplifier of FIG. 1, the magnitude of the load voltage can go up 2Vsat. As a result, the H-bridge amplifier is capable of delivering a peak power to its load four times that of the single-amplifier circuit of FIG. 2a or FIG. 2b. However, two amplifiers need to be powered in the H-bridge amplifier, resulting in twice the power dissipation as for a single-amplifier circuit. (Here, “power dissipation” is the power consumed by an amplifier; it is not the power delivered to the load.)
It would be advantageous for a linear power amplifier to drive a load at a higher peak power than that of a linear single-amplifier power amplifier, and yet have the power efficiency of a linear single-amplifier power amplifier.